If` A_1 and A_2 ` are two AMs between a and b then ` (2A_1 -A_2) (2A_2 - A_1) = `

Options

A. `(a)/(b)`
B. ` (b)/(a)`
C. `ba `
D. ` a^(2)b`

Answer

To solve the problem, we need to find the value of \((2A_1 - A_2)(2A_2 - A_1)\) where \(A_1\) and \(A_2\) are two arithmetic means between \(a\) and \(b\).

### Step-by-Step Solution:

1. **Understanding Arithmetic Means**:
- If \(A_1\) and \(A_2\) are two arithmetic means between \(a\) and \(b\), then we can express them as:
\[ A_1 = a + d \quad \text{and} \quad A_2 = a + 2d \]
where \(d\) is the common difference.

2. **Finding the Common Difference \(d\)**:
- Since \(A_1\) and \(A_2\) are arithmetic means, we know that:
\[ d = \frac{b - a}{n + 1} \]
For two arithmetic means, \(n = 2\), so:
\[ d = \frac{b - a}{3} \]

3. **Substituting \(d\) into \(A_1\) and \(A_2\)**:
- Now substituting \(d\) into the expressions for \(A_1\) and \(A_2\):
\[ A_1 = a + \frac{b - a}{3} = \frac{3a + b - a}{3} = \frac{2a + b}{3} \]
\[ A_2 = a + 2d = a + 2 \cdot \frac{b - a}{3} = a + \frac{2b - 2a}{3} = \frac{3a + 2b - 2a}{3} = \frac{a + 2b}{3} \]

4. **Calculating \(2A_1 - A_2\)**:
- Now, we calculate \(2A_1 - A_2\):
\[ 2A_1 = 2 \cdot \frac{2a + b}{3} = \frac{4a + 2b}{3} \]
\[ 2A_1 - A_2 = \frac{4a + 2b}{3} - \frac{a + 2b}{3} = \frac{4a + 2b - a - 2b}{3} = \frac{3a}{3} = a \]

5. **Calculating \(2A_2 - A_1\)**:
- Next, we calculate \(2A_2 - A_1\):
\[ 2A_2 = 2 \cdot \frac{a + 2b}{3} = \frac{2a + 4b}{3} \]
\[ 2A_2 - A_1 = \frac{2a + 4b}{3} - \frac{2a + b}{3} = \frac{2a + 4b - 2a - b}{3} = \frac{3b}{3} = b \]

6. **Final Calculation**:
- Now we can substitute back into the expression \((2A_1 - A_2)(2A_2 - A_1)\):
\[ (2A_1 - A_2)(2A_2 - A_1) = a \cdot b \]

### Conclusion:
Thus, the value of \((2A_1 - A_2)(2A_2 - A_1)\) is:
\[ \boxed{ab} \]